Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/secr2)

The rewrite relation of the following TRS is considered.

a(b(x1)) → b(b(b(x1))) (1)
a(c(x1)) → c(b(b(x1))) (2)
b(a(x1)) → a(a(a(x1))) (3)
b(c(x1)) → c(c(c(x1))) (4)
a(x1) → x1 (5)
b(x1) → x1 (6)
c(x1) → x1 (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(a(x1)) → b(b(b(x1))) (8)
c(a(x1)) → b(b(c(x1))) (9)
a(b(x1)) → a(a(a(x1))) (10)
c(b(x1)) → c(c(c(x1))) (11)
a(x1) → x1 (5)
b(x1) → x1 (6)
c(x1) → x1 (7)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(a(x1)) → b#(b(b(x1))) (12)
b#(a(x1)) → b#(b(x1)) (13)
b#(a(x1)) → b#(x1) (14)
c#(a(x1)) → b#(b(c(x1))) (15)
c#(a(x1)) → b#(c(x1)) (16)
c#(a(x1)) → c#(x1) (17)
a#(b(x1)) → a#(a(a(x1))) (18)
a#(b(x1)) → a#(a(x1)) (19)
a#(b(x1)) → a#(x1) (20)
c#(b(x1)) → c#(c(c(x1))) (21)
c#(b(x1)) → c#(c(x1)) (22)
c#(b(x1)) → c#(x1) (23)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.